In the answers to Qiaochu's post on defining representations of finite groups over the algebraic integers, it came out that which fields a representation of a finite group is defined over might depend on whether you require your representation to be free or just projective. If your representation V is defined over a number field K, it contains a projective submodule V' for the integers O of K such that V'\otimes_O K=V, but it's not at all clear if this can be chosen to be free.

Luckily, there is a very nice theory of algebraic number theory that says that any projective module over the ring of integers of a number field becomes free when you extend scalars to the Hilbert class field. So, since all representations of finite groups are defined over cyclotomic fields (in fact, you just need the roots of unity for the orders of elements in G), every representation has as an integral basis in the Hilbert class field of a cyclotomic field. Which is.....?

Edit: while the class numbers are interesting that's not the question I asked. I want to actually what the Hilbert class field is, or something about it. For example, is it cyclotomic (seems unlikely, but cyclotomic fields are nice...)?

It's still not obvious to me (although I believe it) why all representations of finite groups are defined over cyclotomic fields. Could you sketch that argument?
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Qiaochu YuanOct 18 '09 at 1:24

My recollection is it's not super easy. You can see the characters are in a cyclotomic extension by noting that all the eigenvalues of a matrix of finite order are roots of unity, but your representation doesn't have to be defined over the field gotten by adjoining character values.
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Ben Webster♦Oct 18 '09 at 1:51

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I believe this is a consequence of Brauer's theorem: this is true for 1-dimensional representations, and every representation's character is an integral linear combination of characters of 1-dimensional representations (by Brauer's theorem and the theorem on representations of supersolvable groups).
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Akhil MathewOct 18 '09 at 2:14

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Ben- if a character of a representation V is an integral linear combination of characters of representations definable over a field k, then V is definable over k (Prop. 33 on p.91, Linear Representations of Finite Groups by Serre). This seems to be how Serre proves the result in question (p.94, Corollary).
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Akhil MathewOct 18 '09 at 3:31

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Oops, I realize I wasn't clear earlier: Each character of G is an integral combination of characters induced from 1-dimensional subgroups. One reduces by the previous comment to the case of 1-dimensional representations, which is clear.
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Akhil MathewOct 18 '09 at 4:00

5 Answers
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Giving an "explicit" description of the Hilbert class field of a number field K (or, more generally, all abelian extensions of K) is Hilbert's 12th problem, and has only been solved for Q and for imaginary quadratic fields. The Hilbert class field H of Q(zeta) will only be contained in a cyclotomic field if H = Q(zeta) itself --- since one can explicitly compute that any abelian extension of Q properly containing Q(zeta) is ramified at some place.

". . .if you fix p, and study the fields K_n obtained by adjoining a p^n-th root of unity to Q, then I believe that the exponent of p in the class number is independent of n (at least for n large enough). . ."

The correct (but still vague) statement here would be, not that the p part of the class numbers are independent of n for large n, but that the growth of the class numbers can be described very explicitly in terms of n. Roughly speaking: the p-part of the class number of K_n has exponent mp^n+ln+v for some integers m, l, v.

How do you add comments below answers by the way? I would have put this under the quoted answer but I couldn't seem to find the right button to push. . .
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Joel DodgeOct 18 '09 at 7:36

I think you need more reputation to do that.
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Qiaochu YuanOct 18 '09 at 7:37

Yes, your general formula is what holds in the case of a general Zp/cyclotomic extension of a number field, but if you take Q as your ground field, then certainly the integer m in your notation is zero (this integer is normally called the mu-invariant). I believe that it is also true that the integer l is zero, and then the p-part would be independent of n for n large enough. Am not sure about l though, the only reason for believing that it is zero was that I've heard of a conjecture of Greenberg saying that l=0 for any totally real field.
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Andreas HolmstromOct 18 '09 at 12:08

Indeed, Greenberg has a conjecture that l=0 if the base field is a totally real field but it is just a conjecture at this point. In the situation under discussion (with K_n = Q(zeta_{p^n})), the base field is Q(zeta_p). Id be surprised to hear that there wasnt a known counterexample to l=0 in this case.
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Joel DodgeOct 18 '09 at 17:34

Ah, then you are right, sorry, I am a bit rusty on cyclotomic fields. I just assumed Q would be the base field in this situtation. It would still be interesting to hear about such a counterexample though.
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Andreas HolmstromOct 18 '09 at 21:42

It follows from class field theory that for any number field K, there is an isomorphism between the class group of K and the Galois group Gal(H/K), where H is the Hilbert class field of K. In particular, the degree of H over K is the class number of K. As David points out, class numbers of cyclotomic fields are complicated, and closely related to classical Iwasawa theory, see for example the book of Washington on cyclotomic fields.

Example: If you take K = Q(zeta) where zeta is a p-th root of unity, p an odd prime, then the class number of K tends to grow with p. In particular, the class number is one iff p < 20.

One the other hand, if you fix p, and study the fields K_n obtained by adjoining a p^n-th root of unity to Q, then I believe that the exponent of p in the class number is independent of n (at least for n large enough), and the exponent of any other prime is bounded as n goes to infinity. Maybe one can say something more precise about these exponents.

Since the Hilbert class field is defined to be the maximal unramified extension of a number field, I think it should be easy to see that the Hilbert class field of a given cyclotomic field is not cyclotomic unless the class number of the base field is 1/